Elasticity/Axially loaded wedge

The BCs at ${\displaystyle \theta =\pm \beta }$ are

${\displaystyle {\text{(30)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}{}=\pm {\widehat {\mathbf {e} }}{\theta }\Rightarrow \sigma _{r\theta }=\sigma _{\theta \theta }=0}$

Assume that ${\displaystyle \sigma _{r\theta }(r,\theta )=0}$. This satisfies the traction BCs on ${\displaystyle \theta =\pm \beta }$ and equation (34). Therefore,

${\displaystyle {\text{(35)}}\qquad \sigma _{r\theta }={\frac {\partial }{\partial }}{}{r}\left({\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }\right)=0\Rightarrow \varphi =r\eta (\theta )+\zeta (r)}$

Hence,

${\displaystyle {\text{(36)}}\qquad \sigma _{\theta \theta }={\frac {\partial ^{2}}{\partial \varphi \partial r}}=\zeta ^{''}(r)}$

That means ${\displaystyle \sigma _{\theta \theta }}$ is independent of ${\displaystyle \theta }$. Therefore, in order to satisfy the BCs, ${\displaystyle \sigma _{\theta \theta }=0}$, i.e.,

${\displaystyle {\text{(37)}}\qquad \zeta (r)=C_{1}r+C_{2}\Rightarrow \varphi =r\eta (\theta )+C_{1}r=r[\eta (\theta )+C_{1}]=r\xi (\theta )}$

Checking for compatibility, ${\displaystyle \nabla ^{4}{\varphi }=0}$, we get

${\displaystyle {\text{(38)}}\qquad \xi ^{(IV)}(\theta )+2\xi ^{''}(\theta )+\xi (\theta )=0}$

The general solution is

${\displaystyle {\text{(39)}}\qquad \xi (\theta )=A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta }$

Therefore,

${\displaystyle {\text{(40)}}\qquad {\varphi =r\left[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta \right]}}$

The only non-zero stress is ${\displaystyle \sigma _{rr}}$.

${\displaystyle {\text{(41)}}\qquad \sigma _{rr}={\frac {1}{r}}\left[2C\cos \theta -2D\sin \theta \right]}$

Plugging into equation (33), we get

${\displaystyle {\text{(42)}}\qquad -D\left[2\beta -\sin(2\beta )\right]=0\Rightarrow D=0}$

Hence,

${\displaystyle {\text{(43)}}\qquad \sigma _{rr}={\frac {2C}{r}}\cos \theta }$

Plugging into equation (32), we get

${\displaystyle {\text{(44)}}\qquad -P=C\left[2\beta +\sin(2\beta )\right]\Rightarrow C={\frac {-P}{2\beta +\sin(2\beta )}}}$

Therefore,

${\displaystyle {\text{(45)}}\qquad {\varphi =Cr\theta \sin \theta ={\frac {-Pr\theta \sin \theta }{2\beta +\sin(2\beta )}}}}$

The stress state is

${\displaystyle {\text{(46)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{r[2\beta +\sin(2\beta )]}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}}$

${\displaystyle {\begin{aligned}2\mu u_{r}&=-{\frac {\partial }{\partial }}{\varphi }{r}+\alpha r{\frac {\partial }{\partial }}{\psi }{\theta }\\2\mu u_{\theta }&=-{\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }+\alpha r^{2}{\frac {\partial }{\partial }}{\psi }{r}\end{aligned}}}$

where

${\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0\\{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\end{aligned}}}$

Plug in ${\displaystyle \varphi =Cr\theta \sin \theta }$,

${\displaystyle {\begin{aligned}&{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\\\Rightarrow &{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)={\frac {2C}{r}}\cos \theta \\\Rightarrow &r{\frac {\partial }{\partial }}{\psi }{\theta }=2C\ln r\cos \theta +A(\theta )\\\Rightarrow &{\frac {\partial }{\partial }}{\psi }{\theta }=2C{\frac {\ln r}{r}}\cos \theta +{\frac {A(\theta )}{r}}\\\Rightarrow &\psi =2C{\frac {\ln r}{r}}\sin \theta +{\frac {\eta (\theta )}{r}}+\xi {r}\end{aligned}}}$

Plug ${\displaystyle \psi }$ into ${\displaystyle \nabla ^{2}{\psi }=0}$,

${\displaystyle {\begin{aligned}&{\frac {1}{r^{3}}}\eta ^{''}(\theta )+{\frac {1}{r^{3}}}\eta (\theta )+\xi ^{''}(r)+{\frac {1}{r}}\xi ^{'}(r)-{\frac {4C}{r^{3}}}\sin \theta =0\\\Rightarrow &\eta ^{''}(\theta )+\eta (\theta )+r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)-4C\sin \theta =0\end{aligned}}}$

Hence,

${\displaystyle {\begin{aligned}\eta ^{''}(\theta )+\eta (\theta )-4C\sin \theta &=b\\r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)&=-{\frac {b}{r^{3}}}\end{aligned}}}$

Solving,

${\displaystyle {\begin{aligned}\eta (\theta )&=-2C\theta \cos \theta +d\cos \theta +e\sin \theta +b\\\xi ^{'}(r)&={\frac {f}{r}}+{\frac {b}{r^{2}}}\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}2\mu u_{r}&=2\alpha C\ln r\cos \theta +(2\alpha -1)C\theta \sin \theta +\alpha (e-2C)\cos \theta -\alpha d\sin \theta \\2\mu u_{\theta }&=-2\alpha C\ln r\sin \theta +(2\alpha -1)C\sin \theta +(2\alpha -1)C\theta \cos \theta -\alpha d\cos \theta -\alpha e\sin \theta +\alpha fr\end{aligned}}}$

To fix the rigid body motion, we set ${\displaystyle u_{\theta }=0}$ when ${\displaystyle \theta =0}$, and set ${\displaystyle u_{r}=0}$ when ${\displaystyle \theta =0}$ and ${\displaystyle r=L}$.Then,

${\displaystyle {\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\cos \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \sin \theta \\u_{\theta }&=-{\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\sin \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \cos \theta -{\frac {C}{2\mu }}\sin \theta \end{aligned}}}$

The displacements are singular at ${\displaystyle r=0}$ and ${\displaystyle r=\infty }$. At ${\displaystyle \theta =0}$,

${\displaystyle {\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\\u_{\theta }&=0\end{aligned}}}$

Is the small strain assumption satisfied ?